package com.wc.算法提高课.F第六章_基础算法.递推与递归.约数之和;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

/**
 * @Author congge
 * @Date 2024/10/1 11:41
 * @description https://www.acwing.com/problem/content/99/
 */
public class Main {
    /**
     * 思路：
     * A = /mul (pi ^ ai)
     * A ^ B = /mul (pi ^ (ai * B))
     * 约束个数:  /mul (1 + ai)
     * 约束之和： /mul (/sum_0_ai pi^j)
     * sum(p, k) = 1 + p + p ^ 2 + ... + p ^ (k - 1)
     * k 为偶数时
     * sum(p, k)
     * = 1 + p + ... + p ^ (k / 2 - 1) + p ^ (k / 2) + ... + p ^ (k - 1)
     * = sum(p, k / 2) + p ^ (k / 2) * sum(p, k / 2)
     * = (1 + p ^ (k / 2) * sum(p, k / 2)
     * <p>
     * k 为奇数时
     * sum(p, k)
     * = 1 + p + ... + p ^ (k / 2 - 1) + p ^ (k / 2) + ... + p ^ (k - 1)
     * = 1 + p + ... + p ^ (k / 2 - 1) + p ^ (k / 2) + ...+ p ^ (k - 2) + p ^ (k - 1)
     * = sum(p, k - 1) + p ^ (k - 1)
     * = (1 + p ^ ((k - 1)/ 2) * sum(p, (k - 1) / 2) + p ^ (k - 1)
     */
    static FastReader sc = new FastReader();
    static PrintWriter out = new PrintWriter(System.out);
    static int a, b, P = 9901;

    public static void main(String[] args) {
        a = sc.nextInt();
        b = sc.nextInt();
        int res = 1;
        for (int i = 2; i * i <= a; i++) {
            if (a % i == 0) {
                int s = 0;
                while (a % i == 0) {
                    a /= i;
                    s++;
                }
                res = res * sum(i, s * b + 1) % P;
            }
        }
        if (a > 1) res = res * sum(a, b + 1) % P;
        if (a == 0) res = 0;
        out.println(res);
        out.flush();
    }

    static int qmi(int a, int b) {
        a %= P;
        int res = 1;
        while (b > 0) {
            if ((b & 1) == 1) res = res * a % P;
            a = a * a % P;
            b >>= 1;
        }

        return res;
    }

    static int sum(int p, int k) {
        if (k == 1) return 1;
        if ((k & 1) == 0) return (1 + qmi(p, k / 2)) * sum(p, k / 2) % P;
        return (sum(p, k - 1) + qmi(p, k - 1)) % P;
    }
}

class FastReader {
    StringTokenizer st;
    BufferedReader br;

    FastReader() {
        br = new BufferedReader(new InputStreamReader(System.in));
    }

    String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

    int nextInt() {
        return Integer.parseInt(next());
    }

    String nextLine() {
        String s = "";
        try {
            s = br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return s;
    }

    long nextLong() {
        return Long.parseLong(next());
    }

    double nextDouble() {
        return Double.parseDouble(next());
    }

    // 是否由下一个
    boolean hasNext() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                String line = br.readLine();
                if (line == null)
                    return false;
                st = new StringTokenizer(line);
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return true;
    }
}
